update existing row in mysql using php coding - php

Question

in my db i have 20+ columns. i added 19 columns throught input form and stored in db succesfully. i fetch few rows from db in my main page. in my main page 1 more column is there. that is status column, it is a combo box type. if i click status column it should show 4 values. i want to select one of the values and then when i click save button it must go to stored in db with that same ID. how to do that? i tried but its not updated in mysql db...
mainpage combo box coding:
echo "\t<td><form action=statusdb.php method=post>
<select name=update><option value=empty></option><option value=Confirm>Confirm</option><option value=Processing>Processing</option><option value=Pending>Pending</option><option value=Cancelled>Cancelled</option></select>
<input name=\"update[".$a_row['slno']."]\"; type=submit id=id value=Save></form>
</td>\n";
status db coding:
if (isset($_GET['id']))
{
$id = mysql_real_escape_string($_GET['id']);
$sql = mysql_query("UPDATE guestdetails SET status = '" . $_POST['update'] ."'");
if(!$sql)
{
die("Error" .mysql_error());
}
}
help me how to do that?

Answer 1

In your IF should be $_POST, not $_GET
Also, need to add WHERE clause, like this:
if (isset($_POST['id']))
{
$id = mysql_real_escape_string($_POST['id']);
$update= mysql_real_escape_string($_POST['update']);
$sql = mysql_query("UPDATE guestdetails SET status = '$update' WHERE id='$id'");
if(!$sql)
{
die("Error" .mysql_error());
}
}
Also, you used name update twice, once in <select> once in <input>, take that one from <input> out, make it hidden field with name id and value of your slno row:
echo "\t<td>
<form action=statusdb.php method=post>
<select name=update>
<option value=empty></option>
<option value=Confirm>Confirm</option>
<option value=Processing>Processing</option>
<option value=Pending>Pending</option>
<option value=Cancelled>Cancelled</option>
</select>
<input name='id' type='hidden' value='".$a_row['slno']."';>
<input type='submit'>Save</button>
</form>
</td>\n";

Answer 2

Try like below:
<form action="statusdb.php" method="post">
<?php
while($a_row = mysql_fetch_array($sql)) {
$sl_no = $a_row['slno'];
echo '<select name="update['.$sl_no.']"><option value=empty></option><option value=Confirm>Confirm</option><option value=Processing>Processing</option><option value=Pending>Pending</option><option value=Cancelled>Cancelled</option></select> ';
echo '<input type="hidden" name="sl_no[]" value="'.$sl_no.'" />';
}
?>
<input name="update_rows" type="submit" value="Save">
</form>
<?php
if(isset($_POST['update_rows'])) {
$nums = $_POST['sl_no'];
$update = $_POST['update'];
foreach($nums as $sl) {
$sl_no = $sl;
$val = $update[$sl_no];
$sql_update = mysql_query("UPDATE guestdetails SET status = '$val' WHERE sl_no='$sl_no'");
}
}
?>

Answer 3

you are not submitting any id to status db.